【DP】 HDOJ 5236 Article

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令dp[i]代表打出i个字符所需的期望，res[i]代表一口气打出i个字符的期望。对于res的转移有res[i] = (res[i-1]+1) * (1-p) + (res[i]+res[i-1]+1) * p。对于dp有dp[i] = min(dp[i], dp[j] + res[i-j] + x)。。由于收敛性，枚举的j不会超过10.

#include <iostream>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 100005 #define maxm 200005#define eps 1e-7#define mod 998244353#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}// headdouble res[maxn];double dp[maxn];int n, x;double p;void work(int _){scanf("%d%lf%d", &n, &p, &x);res[1] = 1.0 / (1.0 - p);int cnt = 200;double mx = (res[1] + x) * n;for(int i = 2; i <= 200; i++) {res[i] = (res[i-1] + 1.0) / (1.0 -  p);if(res[i] > mx) {cnt = i;break;}}dp[1] = res[1] + x;for(int i = 2; i <= n; i++) {dp[i] = dp[i-1] + res[1] + x;if(i <= cnt) dp[i] = min(dp[i], res[i] + x);for(int j = i-1; j > 0 && j >= i - cnt; j--)dp[i] = min(dp[i], dp[j] + res[i - j] + x);}printf("Case #%d: %.6f\n", _, dp[n]);}int main(){int _;scanf("%d", &_);for(int i = 1; i <= _; i++) work(i);return 0;}

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